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Algebra - Graphs in Practical Situations

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Distance-Time graphs represent the relationship between the distance traveled and time taken. The gradient of the graph represents the speed of the object. A horizontal line indicates that the object is stationary (distance remains constant).

A distance-time graph showing constant speed, a stationary period, and then a slower constant speed.
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Speed-Time graphs show how velocity changes over time. The gradient represents acceleration (a=ΔvΔta = \frac{\Delta v}{\Delta t}), where a positive gradient is acceleration and a negative gradient is deceleration. The area under the graph represents the total distance traveled.

A speed-time graph with a shaded area representing distance traveled.
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Real-world linear graphs often take the form y=mx+cy = mx + c, where cc represents the initial value (the yy-intercept) and mm represents the rate of change (gradient). For example, in a phone contract graph, cc would be the fixed monthly fee and mm would be the cost per minute of calls.

A linear graph starting at a y-intercept representing fixed costs.
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Conversion graphs are used to change values between two different units (e.g., Celsius to Fahrenheit or Kilometers to Miles). These are typically straight lines passing through the origin if the units are directly proportional.

📐Formulae

Gradient (m)=y2−y1x2−x1\text{Gradient (m)} = \frac{y_2 - y_1}{x_2 - x_1}

Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

Acceleration (a)=Final Velocity (v)−Initial Velocity (u)Time taken (t)\text{Acceleration (a)} = \frac{\text{Final Velocity (v)} - \text{Initial Velocity (u)}}{\text{Time taken (t)}}

Distance=Area under the Speed-Time Graph\text{Distance} = \text{Area under the Speed-Time Graph}

Area of a Trapezium=12(a+b)h\text{Area of a Trapezium} = \frac{1}{2}(a + b)h

💡Examples

Problem 1:

A car travels from point A to point B. It covers 150150 km in 2.52.5 hours. It then stops for 3030 minutes. Finally, it travels another 6060 km in 11 hour. Calculate the average speed for the entire journey.

Solution:

Total Distance=150+60=210 km\text{Total Distance} = 150 + 60 = 210 \text{ km} Total Time=2.5+0.5+1=4 hours\text{Total Time} = 2.5 + 0.5 + 1 = 4 \text{ hours} Average Speed=2104=52.5 km/h\text{Average Speed} = \frac{210}{4} = 52.5 \text{ km/h}

Explanation:

To find the average speed, we sum all distances and divide by the total time elapsed, including the time the car was stationary.

Problem 2:

A cyclist accelerates from rest to a speed of 1212 m/s in 88 seconds. They then maintain this speed for 2020 seconds before coming to a stop in another 44 seconds with constant deceleration. Calculate the total distance traveled.

Solution:

Area 1 (Triangle)=12×8×12=48 m\text{Area 1 (Triangle)} = \frac{1}{2} \times 8 \times 12 = 48 \text{ m} Area 2 (Rectangle)=20×12=240 m\text{Area 2 (Rectangle)} = 20 \times 12 = 240 \text{ m} Area 3 (Triangle)=12×4×12=24 m\text{Area 3 (Triangle)} = \frac{1}{2} \times 4 \times 12 = 24 \text{ m} Total Distance=48+240+24=312 m\text{Total Distance} = 48 + 240 + 24 = 312 \text{ m}

Explanation:

The distance is the area under the speed-time graph. We divide the area into three parts: the acceleration phase (triangle), the constant speed phase (rectangle), and the deceleration phase (triangle).

Problem 3:

A water tank is being filled. The volume VV in liters after tt minutes is given by the graph of V=15t+100V = 15t + 100. Identify the initial volume and the rate at which the tank is being filled.

Solution:

Comparing with y=mx+c:\text{Comparing with } y = mx + c: c=100  ⟹  Initial Volume=100 litersc = 100 \implies \text{Initial Volume} = 100 \text{ liters} m=15  ⟹  Rate of filling=15 liters/minutem = 15 \implies \text{Rate of filling} = 15 \text{ liters/minute}

Explanation:

In a linear practical graph, the y-intercept represents the starting value (at t=0t = 0) and the gradient represents the constant rate of change.

Problem 4:

A car accelerates from 00 m/s to 2020 m/s in 1010 seconds. It then travels at this constant speed for 2020 seconds before decelerating to a stop in 55 seconds. Calculate the total distance traveled during the journey.

Speed-time graph showing acceleration, constant speed, and deceleration.

Solution:

  1. Total distance is the area under the speed-time graph.
  2. Divide the area into three parts: a triangle (00 to 1010s), a rectangle (1010 to 3030s), and a triangle (3030 to 3535s).
  3. Area 1 (Triangle): 12×10×20=100\frac{1}{2} \times 10 \times 20 = 100 m
  4. Area 2 (Rectangle): 20×20=40020 \times 20 = 400 m
  5. Area 3 (Triangle): 12×5×20=50\frac{1}{2} \times 5 \times 20 = 50 m
  6. Total distance: 100+400+50=550100 + 400 + 50 = 550 m

Explanation:

The distance in a speed-time graph is found by calculating the area of the shape formed between the line and the x-axis. Here, it forms a trapezium, which can be split into simpler shapes for easier calculation.

Problem 5:

A plumber charges a fixed call-out fee plus an hourly rate. The graph of his total charges CC against time hh in hours is shown. Find the call-out fee and the hourly rate.

A cost graph starting at 40 and increasing linearly with time.

Solution:

  1. The call-out fee is the yy-intercept (where h=0h=0). From the graph, at h=0h=0, C=40C = 40. Call-out fee = Rs 40.
  2. The hourly rate is the gradient of the line.
  3. Take two points: (0,40)(0, 40) and (4,120)(4, 120).
  4. Gradient=120−404−0=804=20\text{Gradient} = \frac{120 - 40}{4 - 0} = \frac{80}{4} = 20.
  5. Hourly rate = Rs 20 per hour.

Explanation:

In practical cost graphs, the intercept on the vertical axis represents the fixed or initial cost, while the slope (gradient) represents the variable rate per unit of time or quantity.